Question

The value of $\sec 40^\circ + \sec 80^\circ + \sec 160^\circ$ will be
A. 4
B. -4
C. 6
D. 8

Answer 1

First convert the secant into cosine term by the formula $\sec \theta = \dfrac{1}{{\cos \theta }}$. After that take LCM of the terms. Then find the value of the numerator by applying the formula$\cos A\cos B = \dfrac{1}{2}\left( {\cos \left( {A + B} \right)\cos \left( {A - B} \right)} \right)$. To find the value of the denominator part, multiply and divide the term by $8\sin 40^\circ$ and apply the formula $2\sin A\sin B = \sin 2A$. After that substitute the value and do calculations to get the desired result.

Complete step-by-step solution:
As we know that $\sec \theta = \dfrac{1}{{\cos \theta }}$. Then,
$\Rightarrow \sec 40^\circ + \sec 80^\circ + \sec 160^\circ = \dfrac{1}{{\cos 40^\circ }} + \dfrac{1}{{\cos 80^\circ }} + \dfrac{1}{{\cos 160^\circ }}$
Take LCM of the term,
$\Rightarrow \sec 40^\circ + \sec 80^\circ + \sec 160^\circ = \dfrac{{\cos 80^\circ \cos 160^\circ + \cos 40^\circ \cos 160^\circ + \cos 40^\circ \cos 80^\circ }}{{\cos 40^\circ \cos 80^\circ \cos 160^\circ }}$..............….. (1)
Now, take the numerator part,
$\Rightarrow \cos 80^\circ \cos 160^\circ + \cos 40^\circ \cos 160^\circ + \cos 40^\circ \cos 80^\circ$
We know that,
$\cos A\cos B = \dfrac{1}{2}\left( {\cos \left( {A + B} \right)\cos \left( {A - B} \right)} \right)$
Apply the formula in the above expression,
$\begin{array}{l} \Rightarrow \dfrac{1}{2}\left( {\cos \left( {80^\circ + 160^\circ } \right) + \cos \left( {160^\circ - 80^\circ } \right)} \right) + \dfrac{1}{2}\left( {\cos \left( {40^\circ + 160^\circ } \right) + \cos \left( {160^\circ - 40^\circ } \right)} \right)\\\ \+ \dfrac{1}{2}\left( {\cos \left( {40^\circ + 80^\circ } \right) + \cos \left( {80^\circ - 40^\circ } \right)} \right) \end{array}$
Take $\dfrac{1}{2}$ common from each term,
$\Rightarrow \dfrac{1}{2}\left( {\cos 240^\circ + \cos 80^\circ + \cos 200^\circ + \cos 120^\circ + \cos 120^\circ + \cos 40^\circ } \right)$
Substitute the values,
$\Rightarrow \dfrac{1}{2}\left( { - \dfrac{1}{2} + \cos 80^\circ + \cos 200^\circ - \dfrac{1}{2} - \dfrac{1}{2} + \cos 40^\circ } \right)$
We know that,
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
Applying the formula in the above expression,
$\Rightarrow \dfrac{1}{2}\left( { - \dfrac{3}{2} + \cos 80^\circ + 2\cos \left( {\dfrac{{200^\circ + 40^\circ }}{2}} \right)\cos \left( {\dfrac{{200^\circ - 40^\circ }}{2}} \right)} \right)$
Simplify the terms,
$\Rightarrow \dfrac{1}{2}\left( { - \dfrac{3}{2} + \cos 80^\circ + 2\cos 120^\circ \cos 80^\circ } \right)$
Substitute the value of $\cos 120^\circ$,
$\Rightarrow \dfrac{1}{2}\left( { - \dfrac{3}{2} + \cos 80^\circ + 2 \times - \dfrac{1}{2} \times \cos 80^\circ } \right)$
Cancel out the common factors,
$\Rightarrow \dfrac{1}{2}\left( { - \dfrac{3}{2} + \cos 80^\circ - \cos 80^\circ } \right)$
Subtract the terms,
$\Rightarrow \dfrac{1}{2} \times - \dfrac{3}{2}$
Multiply the terms,
$\Rightarrow - \dfrac{3}{4}$
So, the value of the numerator is $- \dfrac{3}{4}$.
Now, take the denominator part,
$\Rightarrow \cos 40^\circ \cos 80^\circ \cos 160^\circ$
Multiply and divide the term by $8\sin 40^\circ$,
$\Rightarrow \cos 40^\circ \cos 80^\circ \cos 160^\circ \times \dfrac{{8\sin 40^\circ }}{{8\sin 40^\circ }}$
Simplify the terms,
$\Rightarrow \dfrac{{4 \times \left( {2\sin 40^\circ \cos 40^\circ } \right)\cos 80^\circ \cos 160^\circ }}{{8\sin 40^\circ }}$
We know that,
$\sin 2\theta = 2\sin \theta \cos \theta$
Apply the formula in the above expression,
$\Rightarrow \dfrac{{4\sin 80^\circ \cos 80^\circ \cos 160^\circ }}{{8\sin 40^\circ }}$
Simplify the terms,
$\Rightarrow \dfrac{{2 \times \left( {2\sin 80^\circ \cos 80^\circ } \right)\cos 160^\circ }}{{8\sin 40^\circ }}$
Again, apply the above formula,
$\Rightarrow \dfrac{{2\sin 160^\circ \cos 160^\circ }}{{8\sin 40^\circ }}$
Apply the formula again,
$\Rightarrow \dfrac{{\sin 320^\circ }}{{8\sin 40^\circ }}$
As we know,
$\sin \left( {360^\circ - \theta } \right) = - \sin \theta$
Apply in the expression,
$\Rightarrow \dfrac{{ - \sin 40^\circ }}{{8\sin 40^\circ }}$
Cancel out the common terms,
$\Rightarrow - \dfrac{1}{8}$
So, the value of the denominator is $- \dfrac{1}{8}$.
Substitute the value of numerator and denominator in equation (1),
$\Rightarrow \sec 40^\circ + \sec 80^\circ + \sec 160^\circ = \dfrac{{ - \dfrac{3}{4}}}{{ - \dfrac{1}{8}}}$
Simplify the terms,
$\therefore \sec 40^\circ + \sec 80^\circ + \sec 160^\circ = 6$
Thus, the value of $\sec 20^\circ + \sec 40^\circ + \sec 80^\circ$ is 6.

Hence, option (C) is the correct answer.

Note: In case of some random degree angle in trigonometric don’t try to find out the value of that term rather try to solve the problem by manipulating in degree by using different trigonometric identities. One of them used has been mentioned above.